Question

The equilibrium constant Kc for the reaction
P4 {G}↔2P2[G}
is 1.4 at 400 degree Celsius. Suppose that 3 moles of P4(g) and 2 moles of P2(g) are mixed in 2 litre container at 400 degree celsius. What is the value of reaction quotient (Qc)?

Answer 1

Answer : The value of reaction quotient $(Q_{C})$ = 3

Solution : Equilibrium is attained when the rate of forward reaction is equal to the rate of backward reaction.

The relation between $K_{C}$ and $Q_{C}$ is,

$K_{C}$ = $Q_{C}$   (at equilibrium)

$K_{C}$ > $Q_{C}$   (reaction will proceed in Forward direction)

$K_{C}$ < $Q_{C}$   (reaction will proceed in backward  direction)

Given,

Moles of $P_{4}$ = 3 moles

Moles of $P_{2}$ = 2 moles

Volume = 2 L

The given balanced reaction is,

$P_{4}(g)\rightarrow 2P_{2}(g)$

at equilibrium, $K_{C}$ = $Q_{C}$

The expression for $(Q_{C})$ of given reaction is,

$Q_{C}=\frac{[P_{2}]^{2}}{[P_{4}]}$            ........(1)

First, we calculate the concentration of $P_{2}$ and $P_{4}$.

Concentration of $P_{4}$ = $\frac{\text{Moles of} P_{4}}{Volume}$ = $\frac{3}{2}$ = 1.5 mole/L

Concentration of $P_{2}$ = $\frac{\text{Moles of} P_{2}}{Volume}$ = $\frac{2}{2}$ = 1 mole/L

put both the concentration values in above expression (1), we get

$Q_{C}=\frac{[1.5]^{2}}{[1]}$ = 3

The value of $Q_{C}$ > $K_{C}$, this means reaction proceed in backward direction, the product converted into reactant.

Answer 2

Answer:2/3

Explanation:

Given equation->

P4(g) <=> 2P2(g)

We have to find out rhe reaction quotient (Qc) for the reaction when,

moles of P4=3

moles of P2=2

volume = 2litre

Rest of the information given is not of our use :)

So,

by the given number of moles we can find out the concentrations of P4 and P2 by using the relation---->

[concentration]=moles/volume

therefore,

[P4]=3/2

And, [P2]=2/2=1

Now wkt,

Qc=[P2]^2/[P4]

(we have to square the conc. of P2 because its stoichiometric coeff. is 2)

on putting values...

Qc=[1]^2/(3/2)

on solving further,

Qc=2/3

Which is our answer.